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WBJEE · Maths · Functions

Let \(f: R \rightarrow R\) be defined by \(f(x)=x^{2}-\frac{x^{2}}{1+x^{2}}\) for all \(x \in R\). Then,

  1. A \(f\) is one-one but not onto mapping
  2. B \(f\) is onto but not one-one mapping
  3. C \(f\) is both one-one and onto
  4. D \(f\) is neither one-one nor onto
Verified Solution

Answer & Solution

Correct Answer

(D) \(f\) is neither one-one nor onto

Step-by-step Solution

Detailed explanation

We have, \[ \begin{aligned} f(x) &=x^{2}-\frac{x^{2}}{1+x^{2}} \\ f(-x) &=(-x)^{2}-\frac{(-x)^{2}}{1+(-x)^{2}} \\ &=x^{2}-\frac{x^{2}}{1+x^{2}}=f(x) \\ \therefore f(-x)=f(x) & \end{aligned} \] So, \(f(x)\) is many one Again, \(f(x)=x^{2}-\frac{x^{2}}{1+x^{2}}\)…